Question

Suppose that a drug test has a 0.91 probability of succesfully identifying a drug user, but has a 0.06 probability of reporting a false positive. A company drug tests it's employees and 13% of them test positive for drug use.

Let T denote "tests positive for drug use" and D denote "drug user."
The probability 0.91 above refers to P(T|D)
The probability 0.06 above refers to P(T|D^c)
The percentage 13% above refers to P(T)

What is the probability a random employee of this company is a drug user?  
Find:
P(T and Dc)=
P(Tc or Dc)=

Answer 1

Given,

P(T|D) = 0.91

P(T|D^c) = 0.06

P(T) = 0.13

By law of total probability,

P(T) = P(D) P(T|D) + P(D^c) P(T|D^c)

P(T) = P(D) P(T|D) + (1-P(D)) P(T|D^c)

0.13 = P(D) * 0.91 + (1-P(D)) * 0.06

0.13 = (0.91 - 0.06) P(D) + 0.06

P(D) = (0.13 - 0.06) / 0.85 = 0.08235294

Probability a random employee of this company is a drug user is 0.08235294

P(D^c) = 1 - P(D) = 1 - 0.08235294 = 0.9176471

P(T and D^c)=  P(T|D^c) P(D^c) = 0.06 * 0.9176471 = 0.05505883

P(T^c and D^c)=  P(T^c|D^c) P(D^c) = (1 - P(T|D^c)) P(D^c) = (1 - 0.06) * 0.9176471 = 0.8625883

P(T^c or D^c)= P(T^c) + P(D^c) - P(T^c and D^c)

= (1 - P(T)) + (1 - P(D)) - P(T^c and D^c)

= (1 - 0.13) + (1 - 0.08235294) - 0.8625883

= 0.9250588